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(C) For the system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero $(D = 0)$.The de

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$-1$

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(C) For the system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero $(D = 0)$.The determinant is given by:$D = \begin{vmatrix} a^3 & (a+1)^3 & (a+2)^3 \ a & a+1 & a+2 \ 1 & 1 & 1 \end{vmatrix} = 0$Applying column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_2$:$D = \begin{vmatrix} a^3 & (a+1)^3 - a^3 & (a+2)^3 - (a+1)^3 \ a & 1 & 1 \ 1 & 0 & 0 \end{vmatrix} = 0$Expanding along the third row:$1 \cdot \begin{vmatrix} (a+1)^3 - a^3 & (a+2)^3 - (a+1)^3 \ 1 & 1 \end{vmatrix} = 0$$((a+2)^3 - (a+1)^3) - ((a+1)^3 - a^3) = 0$$(a^3 + 6a^2 + 12a + 8 - (a^3 + 3a^2 + 3a + 1)) - (a^3 + 3a^2 + 3a + 1 - a^3) = 0$$(3a^2 + 9a + 7) - (3a^2 + 3a + 1) = 0$$6a + 6 = 0 \Rightarrow a = -1$.

The value of $a$ for which the system of equations $a^3x + (a + 1)^3y + (a + 2)^3 z = 0$,$ax + (a + 1)y + (a + 2)z = 0$,and $x + y + z = 0$ has a non-zero solution is:

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